How to get max of repeated values?

I have a matrix

C =

    0.5000    0.5000    0.3000    1.0000    0.8000    0.3000    0.7000    0.7000    0.3000
    0.4000    0.6000    0.8000    0.6000    0.8000    1.0000    0.8000    1.0000    1.2000

I am looking at the C(2,:) row, everytime there is a repeated instance, I need to take the values from C(1,instance) look at them and max them

The end matrix should look like this

    D =

    0.5000    1.0000    0.8000    0.7000    0.3000
    0.4000    0.6000    0.8000    1.0000    1.2000

Trying to explain better

look at C(2,:)

only 1 value of 0.4, so max of it is 0.5

2 values of 0.6, these are .5 and 1.0, max of these is 1

3 values of 0.8, these are .3 .8 and .7, max of these is .8

so on so forth

Answer 1

[the cyclist]

Here is one way;

% Input
C = [0.5000    0.5000    0.3000    1.0000    0.8000    0.3000    0.7000    0.7000    0.3000
     0.4000    0.6000    0.8000    0.6000    0.8000    1.0000    0.8000    1.0000    1.2000];

% Identify the unique values of the second row of C, along with the index to where
% each of those values appear
[uniqueC,~,indexFromUniqueCBackToAll] = unique(C(2,:));

% For convenience, define the number of unique values
numberUniqueC = numel(uniqueC);

% Preallocate the matrix where the max values are stored
maxRow1 = zeros(1,numberUniqueC);

% For each unique value, in order, find the max of the corresponding values
% in row 1
for nc = 1:numberUniqueC
    indexToThisCValue = (indexFromUniqueCBackToAll==nc);
    maxRow1(nc) = max(C(1,indexToThisCValue));
end

% Append the max values to the unique values to create the output
D = [maxRow1; uniqueC]

D = 2×5

0.5000 1.0000 0.8000 0.7000 0.3000 0.4000 0.6000 0.8000 1.0000 1.2000

[Conner Carriere]

Wow that works perfect! thanks for your quick input, now I am going to try and understand it.

Answer 2

[Paul]

Can use splitapply

C = [0.5000    0.5000    0.3000    1.0000    0.8000    0.3000    0.7000    0.7000    0.3000
     0.4000    0.6000    0.8000    0.6000    0.8000    1.0000    0.8000    1.0000    1.2000];

[G,ID] = findgroups(C(2,:));
D = [splitapply(@max,C(1,:),G) ; ID]

D = 2×5

0.5000 1.0000 0.8000 0.7000 0.3000 0.4000 0.6000 0.8000 1.0000 1.2000

Answer 3

[John D'Errico]

Easy enough. Use unique, then it is just a call to accumarray.

C = [0.5000    0.5000    0.3000    1.0000    0.8000    0.3000    0.7000    0.7000    0.3000
    0.4000    0.6000    0.8000    0.6000    0.8000    1.0000    0.8000    1.0000    1.2000];

% First, match the second row with a set of indices. unique does this.
[C2unique,~,Uind] = unique(C(2,:));
% next, use accumarray to find the group maxima, for each repeated element,
% as identified by Uind
C1max = accumarray(Uind,C(1,:)',[numel(C2unique),1],@max);
Cfinal = [C1max';C2unique]

Cfinal = 2×5

0.5000 1.0000 0.8000 0.7000 0.3000 0.4000 0.6000 0.8000 1.0000 1.2000